[(2006×2009+2007×2008)/2]^2
解析
(n-2)(n-1)n(n+1)+1
= (n-2)(n+1)*(n-1)n +1
= (n*n-n-2)(n*n-n)+1
=[(n*n-n-1)-1][(n*n-n-1)+1]+1
=(n*n-n-1)^2 -1+1
=(n*n-n-1)^2
这里n*n-n-1 = (首位两项乘积+中间两项乘积)/2
= [(n-2)(n+1)+(n-1)n]/2
= [n*n-n-2+n*n-n]/2
= (n*n-n-1)
[(2006×2009+2007×2008)/2]^2
解析
(n-2)(n-1)n(n+1)+1
= (n-2)(n+1)*(n-1)n +1
= (n*n-n-2)(n*n-n)+1
=[(n*n-n-1)-1][(n*n-n-1)+1]+1
=(n*n-n-1)^2 -1+1
=(n*n-n-1)^2
这里n*n-n-1 = (首位两项乘积+中间两项乘积)/2
= [(n-2)(n+1)+(n-1)n]/2
= [n*n-n-2+n*n-n]/2
= (n*n-n-1)